Voltages in Photoelectric Circuits

[MrOriginal]

In the photoelectric effect, if you have a frequency above the threshold frequency, electrons are ejected with greater kinetic energy, but the voltage and current of a circuit using the photoelectric effect is the same. If electrons have a greater Ek, they would have a greater speed, and so if they begin traveling through the circuit’s wire with a greater speed, more electrons would be traveling through per second, so current would be higher, and so would voltage; V= IR. But, as above, this is not the case. Is this because all the additional kinetic energy is used up when the electrons enter the wire?

Then again, if they did enter with more energy, the extent of the conversion of potential energy to kinetic through the wire would still be the same as with electrons of lower kinetic energy, so voltage should be the same. However this conflicts with the idea that the electrons travel faster, leading to a higher current and voltage.

I am quite confused. Please help :)

 

[kuruman]

I am quite confused.

You are.

... but the voltage and current of a circuit using the photoelectric effect is the same.

The same as what? What voltage are you talking about? The relevant voltage in the photoelectric effect is the stopping voltage $V_s$ that is required to stop the ejected photoelectrons to reach the anode. A higher frequency of incident photons means higher kinetic energy of ejected electrons which in turn means higher stopping voltage. The current is related to the number of photons per unit time incident on the photocathode; a higher number of incident photons means a higher number of ejected photoelectrons and, therefore, a higher current.

... so current would be higher, and so would voltage; V= IR.

A photocell does not obey Ohm's law. V = IR is inapplicable here. That's what cuases your confusion.